What is the integral of 2xarcsin(x) ?

The integral of 2xarcsin(x) is x^2arcsin(x)-1/2 arcsin(x)+1/4 sin(2arcsin(x))+C

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integral of 2xarcsin(x)

\int 2 x arcsin (x) d x

x^{2} arcsin (x) - \frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x)) + C

Solution steps

\int 2 x arcsin (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x arcsin (x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} arcsin (x) - \int \frac{x ^{2}}{2 1 - x ^{2}} d x

= 2 (\frac{1}{2} x^{2} arcsin (x) - \int \frac{x ^{2}}{2 1 - x ^{2}} d x)

\int \frac{x ^{2}}{2 1 - x ^{2}} d x = \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

= 2 (\frac{1}{2} x^{2} arcsin (x) - \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))))

Simplify

2 (\frac{1}{2} x^{2} arcsin (x) - \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))) : x^{2} arcsin (x) - \frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x))

= x^{2} arcsin (x) - \frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x))

Add a constant to the solution

= x^{2} arcsin (x) - \frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x)) + C

y^{^{''}} + 2 y^{^{'}} + y = 5 sin (t)

\int_{1}^{27} \frac{4 x ^{\frac{2}{3}} + 1}{4 x ^{\frac{1}{3}}} d x

e^{x}, x^{2} - 1, [- 1, 1]

\int 2 x sin (3 x) d x

y = x^{2} - 13, y = 5 - 3 x

What is the integral of 2xarcsin(x) ?
The integral of 2xarcsin(x) is x^2arcsin(x)-1/2 arcsin(x)+1/4 sin(2arcsin(x))+C